Question

A ball thrown up vertically returns to the thrower after$6\sec $ . Find
(a) The velocity with which it was thrown up,
(b) The maximum height it reaches, and
(c) Its position after $4\sec $

Answer 1

Hint
We use the first equation and the second equation of motion to solve these questions. We take the total time as $t$ and time taken for the ball to reach the maximum height as $\dfrac{t}{2}$. At the time $\dfrac{t}{2}$ , the velocity is taken as zero. The ball is thrown in presence of the earth’s gravity only, so we take acceleration as $g = 9.8m/s$. We use the first equation of motion to find the initial velocity with which it was thrown. Then use the second equation of motion to find the maximum height and the position of the ball after $4\sec $using the initial velocity value found before.
First equation of motion $v = u + at$
Second equation of motion $s = ut + \dfrac{1}{2}a{t^2}$
Where Final velocity is represented by $v$, Initial velocity is represented by $u$, Time is represented by $t$, Distance travelled is represented by $s$, and Acceleration is represented by $a$ which is nothing but gravity $g = 9.8m/s$

Complete step by step answer
We know that when a ball is thrown up it takes equal time to go up till the maximum height and to come down on the ground, so, the total time is taken as $t$.
In this process, the time at which the ball reaches its maximum height is at$\dfrac{t}{2} = \dfrac{6}{2} = 3\sec $, half the time of its total travel.
When the ball is at its maximum height, the velocity of the ball is zero at that position for a fraction of second, taking this velocity as $v = 0$
The ball is thrown upward against the direction of earth’s gravity so acceleration $(a)$= $ - 9.8m/{s^2}$
We now have, At $t = 3\sec $, $v = 0$ and $a = - 9.8m/{s^2}$
Now using these values in the first equation of motion we get the initial velocity with which the ball is thrown,
$ v = u + at $
$\Rightarrow 0 = u + gt $
$\Rightarrow 0 = u + ( - 9.8) \times 3 $
$\Rightarrow u = 29.4m/s $
Hence the ball is thrown with a velocity of $u = 29.4m/s$
Now we use the second equation of motion using the value of initial velocity calculated before to get the maximum height. We know that at the time, $t = 3\sec $, the height is maximum, so,
$s = ut + \dfrac{1}{2}a{t^2} $
$\Rightarrow s = (29.4 \times 3) + (\dfrac{1}{2} \times - 9.8 \times {3^2}) $
$\Rightarrow s = 44.1m $
Hence the maximum height the ball reaches is $44.1m$
We know that the ball reaches its maximum height at $t = 3\sec $ and then starts to fall, so for finding position at $t = 4\sec $ we need to take the instant where the ball reaches its maximum height as the starting point.
So, the initial velocity for the ball $(u) = 0$as velocity is zero at maximum height.
Also, now the ball is coming downwards and is in the same direction as the earth’s gravity, so, $(a) = 9.8m/{s^2}$
We now find the distance travelled by the ball in $1\sec $from the top and subtract this distance from maximum height to find the distance of the ball at the time $4\sec $ from the ground.
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2} $
$\Rightarrow s = 0 + \dfrac{1}{2} \times 9.8 \times {1^2} $
$\Rightarrow s = 4.9m $
The distance at 4 sec from the ground is,$s = $$44.1 - 4.9 = 39.2m$.
Hence at the time, $4\sec $ the position of the ball is $39.2m$ from the ground.

Note
If an object is going against gravity we take the acceleration with a negative sign. We can also say that the position of the ball at $4\sec $ is $4.9m$ from the top but while using these for daily applications we measure from the ground which is why we subtract from the maximum height.